Python触发函数引发异常而不修改它

Question

With the code block:

1  import re
2  import sys
3  import traceback
4 
5  def foo(arg):
6      if isinstance(arg, dict):
7         return arg.get('key', 0)
8      return 0
9  
10 arg = {}
11
12
13
14 try:
15     foo(arg)
16 except Exception:
17     lines = traceback.format_exception(*sys.exc_info())
18     for line in lines:
19         m = re.match(r'  File .*, line (\d+), in foo\n.*\n', line)
20         if m and int(m.group(1)) < 5:
21             print("Foo threw!")
22             sys.exit()
23     raise
24 else:
25     sys.stderr.write("foo did not throw!")

The question is: can I add some code logic between line 10-14, thus make the final code control flow can reach at line 21 , where Foo threw! got printed out ? Attention that, codes between line 1-10 and after line 14 can NOT be modified.

I was tried to write another function to raise an exception, replace foo() with it, it didn't work since the if at line 20, which test against < 5 , "protect" the exception happened before line 5.

Answer 1

No, because these lines:

m = re.match(r'  File .*, line (\d+), in foo\n.*\n', line)
if m and int(m.group(1)) < 5

mean that the error must happen in lines 1-4 which come before foo , so it's not even possible if you add an error to the first line of foo .